of 1 53

Atomic, Planetary, and Biological Systems Described Under A Common Idea

By!

Ian Beardsley!

of 2 53

Important To Point Out

You can speak of the structure of the solar system even though it changes with time. This is

important to understand when I refer to sizes of the Moon and the planets, and their orbital

distances.

The whole object of developing a theory for the way planetary systems form is that they meet the

following criterion: They predict the Titius-Bode rule for the distribution of the planets; the

distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The

distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from

the parent star, the outward pressure gradient from the stellar production of radiation, and the

outward inertial forces as a cloud collapses into a flat disc around the central star. These forces

separate the flat disc into rings, agglomerations of material, each ring from which a different

planet forms at its central distance from the star (it has a thickness). In a theory of planetary

formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of

planets today, which was the distribution of the rings from which the planets formed.

of 3 53

Abstract

This paper describes planetary systems, atomic systems, and biological systems under one

idea, as such it is kind of a theory of everything.!

In Part 1 of this paper I show that the system of measuring time we have today, that divides

the day into 24 hours and the hours into 60 minutes, and the minutes into 60 seconds, that

came to us from the ancient Sumerians, who divided the day into 24 hours, and who gave us

the base 60 counting that resulted in our base unit of time, the second, to have the duration it

has today, was a choice that happens to coincide with my theory that shows that our Earth/

Moon/Sun system and atomic systems have a solution in common of the Schrödinger wave

equation with the same base unit of time of 1 second.!

The ancient Sumerians were the ﬁrst to settle down from wandering, following the herds and

hunting them with stone spearpoint to invent agriculture, architecture, mathematics,

astronomy, and writing. They gave us civilization as we know it today. Since their second came

to us from their system of counting and I have found the second to be a base unit and natural

constant for our planetary system and atomic systems in common, I think it is reasonable to

suggest that perhaps they were given their system of mathematics by ancient Aliens. This is

easy to suggest because their writings say they were given their knowledge by “Those who

came from above.”!

Part 2 I have a theory for the star systems, the planets and their suns, wherein such systems are

solved with the Schrödinger wave equation that is used to solve atomic systems. The result is

that star systems and atomic systems, systems on the macroscales and microscales, are

governed by the same underlying principles. It came to pass that this theory indicated an

overlap with biological systems indicating that biological life could be part of the same idea.

Here I develop that.

Part 3 Our solution of the wave equation for the planets gives the kinetic energy of the Earth

from the mass of the Moon orbiting the Earth, but you could formulate based on the Earth

orbiting the Sun. This yields one second as a base unit as well.

Part 4 We want to find what the wave equation solutions are for Jupiter and Saturn because

they significantly carry the majority of the mass of the solar system and thus should embody

most clearly the dynamics of the wave solution to the Solar System.

Part 5 We want to solve a habitable planetary system in general and apply it to another star

system. We must include the solution of its moon as well, because the Moon of the Earth makes

life optimally possible.

of 4 53

Contents

Part 1: The Sumerians Give Us The Unit of a Second

Which Turns Out To Be A Natural Constant………………………………………5

Part 2: Biological Life Part of a Universal Natural

Process In The Universe……………………………………………………………….20

Part 3: The Solar Formulation For The Solution

Of The Wave Equation………………………………………………………………….30

Part 4: Wave Solutions For Jupiter And Saturn………………………………39

Part 5: Modeling Habitable Star Systems……………………………………….44

Appendix 1………………………………………………………………………………….50

Appendix 2: The Data For Verifying The Equations………………………………52!

of 5 53

Part 1: The Sumerians Give Us The Unit of a Second Which Turns Out To Be A Natural

Constant!

of 6 53

While we can measure time by the completion of the Earth in its orbit around the Sun, 365.25

days, or by the rotation of the Earth once in 24 hours giving us the 24 hour day, we measure

the day by a division of it into 24 equal units we call the hour because the ancient Sumerians

divided the day from sunrise to sunset into 12 hours because they could measure time in the

day by the Sun’s position in the sky as determined by a shadow cast by it on the ground that

moves through 12 divisions in a day, each division called an hour. 12 hours are then added on

for the night from sunset to sunrise, so we have the 24 hour day from sunrise to sunrise, or

sunset to sunset. But why choose 12 divisions which gives us the duration of the hour? I

believe the reason is because 12 is the ﬁrst abundant number. An abundant number means it is

divisible evenly by so many integers preceding it, that their sum is greater than the number

itself. That is,12 is evenly divisible by 1,2,3,4,6 which is greater than 12:!

1+2+3+4+6=16>12!

There is more reason to choose 12 and that is what it is divisible by. Firstly it is divisible by the

two smallest primes 2 and 3, the factors by which no integer can by reduced to further. 2 times

3 is 6 and 12 is evenly divisible by 6 and 6 is of primary importance in mathematics in that the

six-sided regular polygon, the regular hexagon, has its radii equal in lengths to it sides,

something, for one thing, that makes it easy as a tool for computing pi ( ) the ratio of the

circumference of a circle to its diameter. This because inscribed in a circle it approximates the

circle, meaning its perimeter of six, if each side is taken as one, gives its diameter is two,

meaning is close to six divided by two is three. Further twelve is evenly divisible by 4, and the

four-sided square is one of the regular polygons that tessellate meaning the square can tile a

surface without leaving gaps. This makes it suitable for a coordinate system to specify a point

in the plane or space. Such a coordinate system is the ﬁrst we learn of in school, and is the

most used. iIn fact, other coordinate systems are deﬁned in terms of it like spherical systems

where the angles are measured from a set of rectangular axes.!

The other two regular polygons that tesselate are the equilateral triangle and our regular

hexagon.!

The Sumerians use a base 60 (sexagesimal) counting system because 60 is evenly divisible by

so much as well:!

1,2,3,4,5,6,10, 12, 15, 20, 30!

Making it an abundant number as well, it adds up to 108, which is much greater than 60. It is

important to have 5 because in the regular pentagon it determines the golden ratio, phi ( ),

which is recurrent throughout Nature. Life more often meets up with 5-fold symmetry, like a

starﬁsh with ﬁve arms, a human with two legs, two arms, and a head, or ﬁve ﬁngers on each

hand. The physical more often meets up with six-fold symmetry, like in the points on a

snowﬂake. We then have our base unit of one second comes from dividing our day into 24

hours and each of those hours into 60 minutes, and each of those minutes into 60 seconds.

Thus the unit of a second comes to us from the ancient Sumerians, but I ﬁnd the unit of a

second is a natural constant for the the atom and the solar system. !

So we have to ask did ancient aliens give the Sumerians their method of measuring time with

the 24 hour day, and base 60 sexagesimal counting. Indeed, they were the ﬁrst to settle down

from hunting with stone spearpoint and following the herds to invent agriculture, architecture,

mathematics, and writing and their myths say they were given this knowledge from “Those who

came from the above”. !

of 7 53

This second I found is the base unit of the orbital dynamics of the solar system. I found I could

create three operators, one that acts on the radius of the proton to its mass, and the other that

acts on the radius of the Sun to its mass, and one that acts on angular momentum to speed.

is Planck’s constant, is the universal constant of gravitation, is the speed of light, is the

ﬁne structure constant, is the radius of a proton, is the mass of a proton, is the mass

of the Moon, is its radius, and the EarthDay is the rotation period of the Earth:!

They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base

unit of one second.

We say there is a Planck constant for the solar system. We do it with a base unit of one second

because equations 1 and 2 suggest we should . We suggest it is such that it is given by the

standard Planck constant for the atom, , times some constant, , and the Kinetic energy of the

Earth.

Where

6).

Where equation 6 comes from equation 3.

We derive the value of our solar Planck constant

(

6α

4πh

)

⋅

= 1secon d

(

(Ear th Da y)

)

⋅

⊙

= 1secon d

⋅

π r

G m

⋅

= 1secon d

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

π r

G m

C =

⋅

2π r

G m

⋅

18769

299792458

⋅

2π (0.833E − 15)

(6.67408E − 11)(1.67262E − 27)

of 8 53

Now we show that our Planck constant for the solar system gives the base unit of one second for

the quantization. We call our solar Planck constant and find the wavelength for the Moon

which is the ground state for the solar system:

Then wavelength associated with the Moon divided by the speed of light should be 1 second if

our planetary system is quantized in terms of the Moon and a base unit of one second. We have

And we see it is, so we have

The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be

inferred from the solution for an electron around a proton in the a hydrogen atom with the

Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates

1.55976565E 33

⋅

kg ⋅ m

⋅

= kg

= energ y

hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon ds ≈ 1.0secon ds

hC =

(

⋅ m ⋅ s

)

(

⋅

)

(

)(

⋅

)

= second s

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

ℏ

⊙

= (hC )K E

earth

= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s

ℏ

⊙

moon

ℏ

⊙

(2.8314E 33)

(6.67408E − 11)(7.34763E 22kg)

= 3.0281E8m

moon

3.0281E8m

299,792,458m /s

= 1.010secon d s

moon

= 1secon d

of 9 53

10)

Its solution for the atom is

11)

12)

is the energy for an electron orbiting protons and , is the orbital shell for an electron with

protons, the orbital number. I find the solution for the Earth around the Sun utilizes the

Moon around the Earth. This is different than with the atom because planets and moons are not

all the same size and mass like electrons and protons are, and they don’t jump from orbit to

orbit like electrons do. I find that for the Earth around the Sun

13)

14)

is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,

is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is

the orbit number of the Earth which is 3 and is the Planck constant for the solar system.

Instead of having protons, we have the radius of the Sun normalized by the radius of

the Moon. It is has always been an amazing fact the the sizes of the Moon and the Sun are such

that given their orbital distances, the Moon as seen from the Earth perfectly eclipses the Sun.

This becomes part of the theory and we suggest it is a condition for sophisticated planets that

harbor life by writing it:

15)

That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is

about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say

the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon

orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and

extreme cold. Let us now see if our solar Planck constant works…

16)

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

E = −

)

2ℏ

ℏ

Z k

K E

= n

⊙

⋅

2ℏ

⊙

2ℏ

⊙

⋅

⊙

⋅

K E

⊙

ℏ

⊙

planet

moon

≈

star

moon

⊙

6.96E8m

1737400m

= 400.5986

of 10 53

=2.727E36J

The kinetic energy of the Earth is

The accuracy of our equation is:

Which is very good. Thus we have solved the Earth/Moon/Sun System with our spacetime

operators by using them to find a Planck constant for the solar system.

We want to suggest that the base 60 dynamic applied to the rotational velocity of the Earth to

measure time has a natural property associated with the atom and the Earth/Moon/Sun orbital

system as a solution to the atom’s wave equation. The Earth day is given by, in seconds

This gives since

Thus we have as factors for the seconds per day the smallest primes 2 and 3, and the 4 of

rectangular coordinate systems, and the versatile, abundant, 60. We then suggest the second is

dynamic in terms of what the ancients gave us because 86400 seconds comes from

and this I suggest is connected to Nature because the Earth rotates at a speed

that gives from these ancient factors the duration of a second that I found is the base unit of the

atomic systems and our planetary system, in particular of our Earth/Moon/Sun system as a

solution of the Schrödinger wave equation that solves the same atomic system, which I will go

into right now.

We see the spacetime operators solve the atom by giving us the radius of a proton. We set

equation 3 equal to equation 1

1).

These two yield

K E

= (1.732)(400.5986)

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

K E

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.727E 36J

2.7396E 33J

100 = 99.5 %

(

1d ay

24hrs

)(

1hr

60min

)(

1min

60sec

)

86400

d ay

sec

2 ⋅ 3 ⋅ 4 = 24

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60 = 86400second s /d a y

2 ⋅ 3 ⋅ 4 ⋅ 60 ⋅ 60

⋅

π r

G m

⋅

= 1secon d

(

6α

4πh

)

⋅

= 1secon d

of 11 53

17).

I find this is close to the experimental value of the radius of a proton. I find I can arrive at this

radius of a proton another way, energy is given by Planck’s constant and frequency

We have

We take the rest energy of the mass of a proton :

The frequency of a proton is

Since our theory gave us the factor of 2/3 for the radius of a proton we have:

The radius of a proton is then

This is close to the CODATA value for the proton radius around 2018 (0.842E-15m). The result

We find it may be that the radius of a proton is actually

18)

⋅

E = h f

f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J

E = J = J oules = energ y

E = m

⋅

≈ ϕ =

⋅

6.62607E − 34

(299,792,458)(1.67262E − 27)

= 0.88094E − 15m

= ϕ ⋅

= 0.816632E − 15m

of 12 53

Where is the golden ratio constant (0.618…). This is more along the lines of more recent

measurements. Both equations 17 and 18 for the radius of the proton can be right depending on

the dynamics of what is going on; the radius of a proton is not precisely defined, it is more of a

fuzzy cloud of subatomic particles. Thus we have solved the atom with our spacetime operators

by producing the radius of a proton.

Indeed if the dynamics of the factors the ancients gave us to create the duration of a second are

connected to the dynamics of stars systems then then the second should define the rotational

angular momentum of the Earth since we divide the rotation period of the Earth into these

factors to get the unit of second, and should be connected to our Planck constant for the solar

system which is in units of angular momentum as well, as is Planck’s constant for the atom.

The angular momentum of the Earth with respect to the Sun is 2.66E40 kg m2/s. The rotational

angular momentum is 7.05E33 kg m2/s. In orbit angular momentum is given by

For a uniform rotational sphere it is given by

We found our solar system Planck constant was

This gives

19)

We are now equipped to show that the ancient Sumerians were right in dividing the rotation

period of the Earth (the day) into 24 units (the hour) because

That is

20).

Which is to say the angular momentum of the Earth to its Planck constant gives the base 60

counting in terms of the 24 hour day, the 60 of 60 minutes in an hour, and 60 seconds in a

minute, that determine our base unit of duration we call a second that happens to be, as I have

shown, the base unit of the wave solution to the atom and the Earth/Moon/Sun system.

Our equation in this paper for the Earth energy as a solution of the wave equation (eq. 13)

13).

L = 2π M f r

L =

π M f r

ℏ

⊙

= 2.8314E 33J ⋅ s

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

2.5(24) = 60

earth

ℏ

⊙

24 = 60

K E

= n

⊙

⋅

2ℏ

⊙

of 13 53

does not depend on the Moon’s distance from the Earth, only its mass. The Moon slows the

Earth rotation and this in turn expands the Moon’s orbit, so it is getting larger, the Earth loses

energy to the Moon. The Earth day gets longer by 0.0067 hours per million years, and the

Moon’s orbit gets 3.78 cm larger per year. Equation 20

20)

only specifies divide the day into 24 units, and hours into 60 minutes and minutes into 60

seconds, regardless of what the Earth rotational velocity is. But it was more or less the same as it

is now when the Sumerians started civilization. But it may be that it holds for when the Earth

day is such that when the Moon perfectly eclipses the Sun, which we said might be a condition

for the optimization of life preventing extreme hot and cold. That is when 15 holds

15)

Which holds for today and held for the ancient Sumerians and holds for when the Earth rotation

gives the duration of a second we have today.

We want the basis set of equations for the solar system. We have

We have from equation

We have equations

earth

ℏ

⊙

24 = 60

planet

moon

≈

star

moon

ℏ

⊙

= 3.0281E8m

moon

ℏ

⊙

⋅

= 1.0secon d s

ℏ

⊙

= (hC )K E

hC = 1secon d

⋅

π r

G m

⋅

= 1secon d

(

6α

4πh

)

⋅

= 1secon d

1secon d =

K E

(Ear th Day)

of 14 53

From these it becomes clear that

21)

22)

23)

Thus combining equation 20 with the following…

21).

20).

Which gives

24)

This is very accurate to give us a second. But notice 6.262 is approximately . We see that

25)

is the circumference of a unit circle, it could be that the circumference of the Earth orbit, can

be taken as 1 (unity) and essentially we have the mystery of base 60 and the 24 hour day of the

ancient Sumerians is solved, it connects unit circle to 1 (unity).

ℏ

⊙

K E

ℏ

⊙

ℏ

⊙

(

6α

⋅

4πh

)

K E

moon

= 1secon d

hC = 1secon d

ℏ

⊙

= (hC )K E

K E

ℏ

⊙

earth

ℏ

⊙

24 = 60

1secon d =

(

)

earth

(

)

(7.05E 33)

(6.67408E − 11)((7.347673E 22kg)

(299792458m /s)

(

)

6.262sec = 1.002secon d s = 1.00secon d s

2π

1 =

(

)

2π

of 15 53

26).

27)

I came up with a physical theory for certain aspects of reality that showed that our solar system

had a wave equation solution, where the wave equation was developed for describing atomic

systems, among other things, and is at the heart of quantum physics, the modern physics that

describes things on the micro-scale and our planetary system is on the macro-scale. I found it

was based on the Moon that orbits the Earth, which seems to be some kind of a natural

yardstick, and was further based on the unit of a second for measuring time as the base unit,

which I found turns out to be some kind of a natural constant. This natural constant of a

second also, in the theory, predicts the radius of a proton, the fundamental unit of the atom,

and as such a gap is bridged between the microcosmos and the macrocosmos. Further it

resulted in a description of the hydrocarbons, which are the backbones of the chemistry of life.

Bringing in the biological like this makes it the rudiments of a theory of everything,!

The interesting thing is, though, that the second was not designed, originally, as a natural

constant, but came to us ultimately from the ancient Sumerians of Mesopotamia, who were

some of the ﬁrst to settle down from hunting with stone spearpoints to invent agriculture,

mathematics, writing, metallurgy, architecture… to invent civilization.!

The idea was to divide the day into twelve hours, and measure each hour by the position of the

Sun in the sky, as measured by the shadow cast by a rod onto a plate with lines to measure the

time of day. They chose twelve hours in a day because 12 is the smallest abundant number,

which is to say it is evenly divisible by 1,2,3,4,6 whose sum is 16 which is greater than twelve. !

Their counting system was base 60 (sexagesimal) which means they started counting again

after counting to 60 where our system is base 10 (decimal) which means we start counting

again after 10. They chose base 60 because sixty, aside from being an abundant number — its

divisors add up to a value greater than it — it is the ﬁrst number divisible by the ﬁrst 6 counting

numbers, 1,2,3,4,5,6 aside from being evenly divisible by 10,12,15,30.!

This lead to the ancient Babylonians dividing the hour into 60 minutes, and the minute into 60

seconds. This in the end lead to the ancient Greeks using it for astronomy, and the Italians to in

the Middle Ages build the ﬁrst clocks with hours, minutes, and, seconds. In the end they gave

us our clocks and they way we measure time in the West, today. The ancient Egyptians,

separately and independently invented civilization from the Sumerians, but were never invaded

by tribes, like in Sumer, who stayed and adopted their math and writing, but rather were

isolated by desert, ocean, and forests, so as not to spread their invention of Mathematics,

writing, and architecture to the rest of the world, at least until they and the ancient Greeks

interacted and shared knowledges in several exchanges where ancient Greek scholars

wandered into Egypt. But, the ancient Egyptians also divided the day into 12 hours and

invented sun dials to tell time as such. However, we got the 12 hour day, or 24 hour day from

sunrise to sunrise and sunset to sunset, from the ancient Sumerians.!

However, myself, having been born in the West I was more aware of Sumer and Egypt, than

India, and indeed while Sumer and Egypt started civilizations, so did the people who brought it

to India, to the Dravidians, and the idea of transmigration of the soul who were from its

outskirts, who brought it from the Indus Valley where the rivers were, they were the Indus Valley

cos(45

∘

) = cos(π /4) =

of 16 53

Civilization, just as old as Sumer and Egypt. It would seem civilizations always started along

rivers, the Tigris and Euphrates in Mesopotamia, and the Nile in Egypt.!

However, after having looked at my theory for reality in terms of the base 60 of Sumer that gave

us the second, and how its dynamics could have coincided with modern physics, and

suggesting that it was given to them by ancient Aliens, because they said they got their

knowledge from those who came from the heavens, the Anunaki, I began to look at the system

that ancient India used to measure time, and I found that they used base 60 as well, but in a

very reversed way from what we got from Mesopotamia, the Babylonians, and ancient Greeks.

And, I found further, that their system of base 60 describes certain aspects of Nature as well.!

Before the colonization of India by the British, another system of measuring time from ancient

times was used, and it was called the Bharat Clock. There are eﬀorts in India today to return to

this, and they suggest for contemporary physics it serves better, as well as it is more eﬃcient

for productivity in the work force. Further still they explain that it is based on Natural

phenomenon in the Universe, and that in the method of the West, it is not. I would take issue

with that not just because my theory for reality shows the Mesopotamian system gives us the

unit of a second in planetary systems, atomic systems, and biological systems, which they

could not know about, but because it does have an elegance that works well for the the

periods of celestial motions, and has since its inception, though I can see advantages in their

system as well.!

The Hindu system is quite inverted from ours, but my theory of physical reality shows the two

systems are profoundly connected in a way that can be quite revealing. We measure the year

as 365.25 days which is close to the 360 degrees in a circle. As such the Earth moves around

the Sun in its orbit through neatly about 1 degree per day. The Sumerians gave us the 360

degrees circle, because in their base 60 counting 6 times 60 is 360, and 6 is the most dynamic

number, a six-sided regular polygon has its radii equal to its sides, which mathematically has

many advantages for solving problems. It occurs in Nature as the closest packing of equal

radius spheres, the so so called six around one, typical in Nature of petal arrangements around

a center in a ﬂower, and 6-fold symmetry is typical of the physical, like snowﬂakes forming six-

points as ice crystals fall through cold air. 5-fold symmetry is typical of life, like two eyes and

nose and a mouth, or two arms, two legs, and a head.!

The Earth then rotates close to 360 times in a year, close to the time for it to complete an orbit

around the Sun. Each rotation is a day, and from the Sumerians we divided that day into 24

hours. 60 divided by 24 is 2.5 and as it would turn out 2.5 is the factor than connects the Hindu

system to the Western system. As we shall see, my theory ﬁnds this factor is a pivotal factor of

Nature. Let’s look at our system and theirs. We have!

(365.25 days)(24 hours)(60 minutes)(60 seconds)=31557600 seconds/year!

And the seconds in a day are!

(24 hours)(60 minutes)(60 seconds)=86400 seconds/day!

Thus we divide the hour into 60 minutes, and the minutes into 60 seconds. The Hindu Vedic

System has a day is 60 ghatika of 24 minutes each, each ghatika is divided into 60 palas of 24

seconds each, and each pala is divided into 60 vispalas, each vispala of 0.4 seconds each. So

where our system has a base unit of 1 second, theirs has a base unit of 0.4 seconds, so that

could be and advantage to their system, a smaller unit of time is more reﬁned. Further their day

is divided into 60 units, ours into only 24, so their hour, the ghatika is only 24 minutes long, and

ours is 60 minutes long, The proponents in India say since when working and doing chores we

of 17 53

do a few chores in an hour, they do about one per ghatika is 24 minutes, which makes the

measure of time more manageable. That could be an advantage I think. They say our hour is so

long because the lines had to be far apart on the Egyptian Sun Dial so the shadow cast by the

Sun didn’t cross-over onto another line. But today, with modern technology we can make clock

lines marking hours closer together and measure them with a pointer hand pointing to them

without any problems, and the result is we would have a smaller more reﬁned hour (60 of them

in a day as opposed to 24). They suggest this method of measuring time would work better in

science and engineering as well, that we have to get away from the way sun dials had to be

made in Egypt in ancient times.!

But they further point out that their system describes Nature. They say theirs are 108,000

vispalas in a day, and 108,000 vispalas in a night giving 216,000 vispalas in a 24 hour day. The

diameter of the Sun is 108 that of the Earth, and the average distance from the Sun to the

Earth is 108 solar diameters, and the average distance from the Moon to Earth is 108 lunar

diameters. 108(10)(10)(10)=108,000. Ourselves and them use base 10 counting, and that is

probably because we have ten ﬁngers to count on.!

Conclusion

In conclusion we have that the unit of a second as derived from the Mesopotamians is a Natural

constant. As well we see the Hindu Vedic system describes Nature as well. It says

( )

To see how this is connected to nature we would have to go into the significance of base 10

counting as we did with base 60 counting. But interesting here is that

⊙

= 2(6.957E8m) = 1.3914E 9m eters

⊙

= SolarRadiu s

= 1.496E11m

= Ear thOrbit

⊙

= 107.52 ≈ 108

moon

= 2(1.74E6m) = 3.48E6m

moon

= Lun arRa dius

moon

= 3.845E8m

moon

= Lun arOrbit

moon

= 110 ≈ 108

(86,000sec /d a y)

43200

0.4sec /vispa l a

= 108,000vispalas /d a y

108(10)(10)(10) = 108,000

⊙

= 107.52 ≈ 108

moon

= 110 ≈ 108

of 18 53

because

13.

where

15.

16.

Which is to say radius of the Sun to the radius of the Moon in the planetary equation plays the

role of the number of protons (Z) in an element for the energy of the electron orbits in the

atomic equation and that the Moon perfectly eclipses the Sun as see from the earth (eq 15) might

be a condition for sophisticated habitable star systems, we know the Moon allows for the

seasons making life very successful on Earth by preventing temperature extremes. So in the

Hindu Vedic system the diameters of the Sun and Moon fitting into Earth orbit and Lunar orbit

the same amount might play a similar role in a theory. You would have to go into base 10 like we

did with base 60. We have

West

24 hours

60 minutes

60 seconds

India

60 hours

24 minutes

24 seconds

A sort of inversion of things. The incredible thing here is that the wonderful equations,

equations 19 and 20:

19.

20.

unify the Mesopotamian system with the Hindu system because

⊙

6.957E8m

6,371,000m

= 109 ≈ 108

K E

= n

⊙

⋅

2ℏ

⊙

planet

moon

≈

star

moon

⊙

6.96E8m

1737400m

= 400.5986

earth

ℏ

⊙

7.05E 33

2.8314E 33

= 2.4899 ≈ 2.5 = 2

earth

ℏ

⊙

24 = 60

of 19 53

This suggests that the Hindu system is just as Natural as the Western system.

Equation 25 that says

25)

1 hour=(60min/hour)(60sec/min)=3600sec=1 Western Hour

1 vispala=0.4sec

60 vispala=(60)(0.4sec)=24 sec=1pala

60 pala = (60)(24 sec)=1440 sec=24 min = 1 ghaticka = 1 HinduHour

28).

where is the circumference of a unit circle, and we have unity on the left, so it could be if you

combine the Indian system of measuring time with the Western system you would have a system

of units for the solar system where the equations work out nicely.

1vi spala = 0.4secon d s

(0.4s)(2.5) = 1secon d

1 =

(

)

2π

1 =

(1440sec)

(3600sec)

2π

1 =

(1gh at ick a)

(1h our)

2π

1 =

Hin du Hour

Wester n Hour

2π

1 =

(60pala)

(60min)

2π =

(24min)

(60min)

2π

1 =

((60)(24sec))

((60)(60sec))

2π =

(60pala)

(60sec)

2π =

(24sec)

(60sec)

2π

of 20 53

Part 2: Biological Life Part of a Universal Natural Process In The Universe

of 21 53

Introduction I have a theory for the star systems, the planets and their suns, wherein such

systems are solved with the Schrödinger wave equation that is used to solve atomic systems. The

result is that star systems and atomic systems, systems on the macro scales and micro scales, are

governed by the same underlying principles. It came to pass that this theory indicated an

overlap with biological systems indicating that biological life could be part of the same idea, and

that is what I hope to pull out of that paper and develop here.

The theory for the planets, which also predicted the radius of a proton, which is one of the

fundamental particles out of which matter is made, suggested that star systems which support

life are a part of a Universal natural process. One of the conditions for star systems that support

life made use of the interesting fact that has mystified science for some time now, that the Moon

is the right size and distance from the Earth that it nearly perfectly eclipses the Sun. The reason

a moon would be necessary for life to be optimally successful on a habitable planet is that we

know our moon that orbits our Earth, makes life here successful because it holds the Earth at its

inclination to its orbit around the Sun allowing for the seasons, thus preventing extreme hot and

extreme cold.

Interestingly, the solution for the planets of the Schrödinger wave equation, was quantized in

terms of the Earth’s moon, and a base unit of one second. The Moon seems to be some sort of a

natural yardstick. The strange thing is that the basis unit of time is one second and the second

was not developed with the planets and atoms in mind, so it is strange that it lands upon the

function of a natural constant in our theory. In the theory (An Abstract Theory of Reality,

Beardsley 2024) I talk a little about why that might be. The second came to us from the Ancient

Sumerians, who invented civilization with settling down from following the herds and hunting

with stone spearpoint to invent agriculture, metallurgy, writing, and mathematics. They had a

base 60 counting system, and found it convenient then to divide the Earth day (rotation period)

into 24 hours, which in the end became adopted by the world. Their base 60 counting resulted in

the Babylonians dividing the hour into 60 minutes, and that in turn into 60 seconds, and that is

how we have the duration of a second we have today. They chose base 60 (sexagesimal) because

60 is evenly divisible by so much from which they gave us the 12 hour day and 12 hour night or

24 hour day from sunrise to sunrise, sunset to sunset. 12 times 5 is 60, also 60 times 6 is 360,

they gave us the 360 degree circle as well.

In this paper I show that the same unit of a second that describes planetary systems and atomic

system in common describes hydrocarbons, the skeletons of biological life chemistry. I further

show that it predicts the atomic radii of the hydrogen and carbon atoms from which such

skeletons are made. Hydrogen and carbon are the most abundant elements in life chemistry,

carbon is the core element upon which life is made, and hydrogen is the simplest element,

element one in the periodic table of the elements, consisting of 1 proton, and is the most

abundant element in the universe by far, and plays the dominant role in all of chemistry.

of 22 53

A Theory For Biological Hydrocarbons I have found that the basis unit of one second is

not just a Natural constant for physical systems like the atom and the planets around the Sun

(See my paper An Abstract Theory of Reality, Beardsley 2024) but for the basis of biological

life, that it is in the sixfold Nature of the chemical skeletons from which life is built, the

hydrocarbons. I found

1).

2).

Where is the fine structure constant, is the radius of a proton, is the mass of a

proton, is Planck’s constant, is the constant of gravitation, is the speed of light, is the

kinetic energy of the Moon, is the kinetic energy of the Earth, and is the

rotation period of the Earth is one day.

The first can be written:

3).

4).

From which instead of saying the left sides of these equations are seconds, we say they are

proton-seconds by not letting the units of cancel with the bodies of these equations on the

left, which are in units of mass, but rather divide into them, giving us a number of protons. I say

this is the biological because these are the hydrocarbons the backbones of biological chemistry.

We see they display sixfold symmetry. I can generate integer numbers of protons from the time

values from these equations for all of the elements with a computer program. Some results are:

1secon d =

6α

⋅

4πh

1secon d =

K E

(Ear th Day)

α = 1/137

K E

Ear th Day

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

of 23 53

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of

the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons

the backbones of biological chemistry. And carbon is the core element of life. We see the

duration of the base unit of measuring time, 1 second, given to us by the ancients (the base 60,

sexagesimal, system of counting of the Sumerians who invented math and writing and started

civilization), is perfect for the mathematical formulation of life chemistry. Here is the code for

the program, it finds integer solutions for time values, incremented by the program at the

discretion of the user:

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792458,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

float decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We have that

Since this 6 seconds is also proton-seconds we have

5). is carbon (C)

6). is hydrogen (H)

⋅

4πh

G c

= 18769

0.833E − 15

1.67262E − 27

4π (6.62607E − 34)

(6.67408E − 11)(299,792458)

= 6.029978s ≈ 6s

6pr ot on s

⋅

4πh

G c

= 1secon d

1pr ot on

⋅

4πh

G c

= 6secon d s

of 24 53

Our Theory For Hydrocarbons With all that has been said we are equipped to proceed. We

want to consider the radius of a hydrogen atom and the radius of a carbon atom. The radius of

a carbon atom given in your periodic table of the elements is often 70 to 76 picometers. The

covalent radius of hydrogen is given as 31 picometers. The atomic radius of hydrogen is 53

picometers and the atomic radius of carbon is 67 picometers. We want to consider the atomic

radii of both, because the covalent radius, determined by x-ray diﬀraction for diatomic

hydrogen, is the size of two hydrogen atoms joined H2 divided by two, where it is measured

that way, joined, in the laboratory. Carbon is C2 divided by 2. We are interested in the single

carbon and hydrogen atoms, because we want to know what our theory for their six-fold

symmetry with one another in their representations in proton-seconds says about the way they

combine as the skeletons of life chemistry. We start with the Planck constant, which is like

ﬂux, a mass (perhaps a number of particles) per second over an area. That is it is kilograms per

second over square area:!

We have equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

We can write these

3).

4). !

We have from 3 and 4…!

5). !

We ﬁnd the ratio between the surface areas of the hydrogen and carbon atoms:!

h = 6.62607E − 34J ⋅ s = 6.62607E − 37

⋅ m

6pr ot on s

⋅

4πh

G c

= 1secon d

1pr ot on

⋅

4πh

G c

= 6secon d s

(

6.62607E − 37

⋅ m

)

6secon d s

(

6.62607E − 37

⋅ m

)

6seconds

1.67262E − 27kg

= 2.37689E − 6m

(

6.62607E − 37

⋅ m

)

1second

(

6.62607E − 37

⋅ m

)

1second

6(1.67262E − 27kg)

= 6.602486E − 8m

⋅

(6seconds)

⋅

(1second)

2.37689E − 6m

6.602486E − 8m

= 35.9999 ≈ 40

of 25 53

6). !

7). !

It is the golden mean. These atomic radii are the radii between the nucleus of the atoms and

their valence shell, which is what we want because the valence shell is the outermost electrons

responsible for the way the hydrogen and the carbon combine to make hydrocarbons. We will

write this!

8). !

I am guessing the reason we have the golden mean here is that it is the number used for

closest packing. But what we really want to do is look at the concept of action, for hydrogen

given by six seconds and carbon given by 1 second. We take equations 1 and 2:!

1). is carbon (C)

2). is hydrogen (H)

And, we write

9).

Where is the radius of the atom, and t its time values given here by equations 1 and 2. We have for

hydrogen

10).

This is actually very close to the radius of a hydrogen which can vary around this depending on how you

are looking at it, which we said is given by 5.3E-11m. For carbon we have:

11).

And this is actually very close to the radius of a carbon atom which is 6.7E-11m. The thing is if we consider

the bond length of the simplest hydrocarbon CH4, methane, which can be thought of as

surface

= 4π (r

) = 3.52989E − 20m

surface

= 4π (r

) = 5.64104E − 20m

4π(53pm)

4π(67pm)

= 0.62575 ≈ 0.618... = ϕ

= ϕ

6pr ot on s

⋅

4πh

G c

= 1secon d

1pr ot on

⋅

4πh

G c

= 6secon d s

G c

∫

dt = r

G c

∫

dt = (1.67262E − 27k g)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27k g)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

of 26 53

16).

our, equations give

17).

which are the same thing. Thus we have the basis for a theory of everything in that it includes the macro

scale, the Earth/Moon/Sun System, because we had:

and it includes the radius of the proton which is the radius and mass of a proton and gives 1

second

and we have the hydrocarbons the skeleton of the chemistry of life give one second

1). is carbon (C)

2). is hydrogen (H)

and because they predict the radii of the carbon and hydrogen atoms at the core of life

10).

11).

We want to proceed with the elements nitrogen (N) and oxygen (O). We just did carbon and hydrogen, but

hydrocarbons are then combined with primarily nitrogen and oxygen to make amino acids, the building

blocks of life. DNA synthesizes them into proteins, for hair, muscles tissue, skin, nails, etc… The most

basic organic compound, HNCO (isocyanic acid) is just one each of these elements. It bonds

H-N=C=O

It was shown by the Miller-Urey experiment that if a primordial Earth was mainly ammonia (NH3),

Methane (CH4), water (H2O) these combined under energy form 11 of the 20 biological amino acids life

needs for DNA to synthesize proteins. Thus we start with nitrogen and notice the first eight element are

atomic number 1 is hydrogen, atomic number 2 is helium, atomic number 3 is lithium, atomic number 4 is

beryllium, atomic number 5 is boron, atomic number six is carbon, atomic number 7 is nitrogen, atomic

number 8 is oxygen. The atomic numbers correspond to the the number of protons in each element.

+ r

= 5.3E − 11m + 6.7E − 11m = 1.2E − 10m

+ r

= 5.5E − 11m + 6.44E − 11m = 1.2E − 10m

1secon d =

K E

(Ear th Day)

1secon d =

6α

⋅

4πh

6pr ot on s

⋅

4πh

G c

= 1secon d

1pr ot on

⋅

4πh

G c

= 6secon d s

G c

∫

dt = (1.67262E − 27k g)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27k g)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

of 27 53

We suggest the periodic table of the elements has its function in sixfold symmetry because it has

18 groups in that 6 times 3 is 18 and 3 this the first odd prime number other than one, where all

prime numbers are odd except 2, and 1 defines a prime number in that a prime number is a

number that is evenly divisible by itself and 1. The periodic table is then centered around

hydrogen and carbon, hydrogen with one proton, carbon with six where hydrogen is given by 6

seconds and carbon is given by 1 second, all values beyond 6 seconds are fractional protons and

1 second is the smallest integer number of seconds that gives whole protons. That is, equations 1

and 2 are

1). is carbon (C)

2). is hydrogen (H)

We already had that hydrogen, the base element with 1 proton gives itself, hydrogen, and carbon with 6

protons is given by itself and hydrogen in equations 10 and 11:

10).

11).

We had that experimentally for the hydrocarbon backbones of life

16).

And, according to our theory that

17).

We want to consider that elements 6, 7, 8 which are carbon, nitrogen, and oxygen, respectively then

correspond to H, He, Li, and Be element 1, 2, 3, and 4. Carbon (6) already corresponds to hydrogen (H),

so nitrogen (7) corresponds to H, He, Li, and oxygen (8) corresponds to H, He, Be. This also works

because nitrogen (group 15) wants to have stable noble gas electron configuration (group 18) and 18-15=3

meaning it ionizes and oxygen (group 16) wants the same so 18-16=4 meaning it oxidizes . Thus

6pr ot on s

⋅

4πh

G c

= 1secon d

1pr ot on

⋅

4πh

G c

= 6secon d s

G c

∫

dt = (1.67262E − 27k g)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec

(9.2E − 12m /s)(6secon d s) = 5.5E − 11m

G c

∫

dt = (1.67262E − 27k g)

(6.67408E − 11)(299,792,458)

6.62607E − 34

∫

6sec+1sec

(9.2E − 12m /s)(7secon d s) = 6.44E − 11m

+ r

= 5.3E − 11m + 6.7E − 11m = 1.2E − 10m

+ r

= 5.5E − 11m + 6.44E − 11m = 1.2E − 10m

2−

of 28 53

lithium is 3 protons, and He+H=3 protons for nitrogen ( ) and oxygen ( ) is Be-He=2. We have

that

We have

Nitrogen

Oxygen

Thus from equation 9

Nitrogen

Where experimentally

And, theoretically

So we are in agreement.

3−

2−

6pr ot on s

⋅

4πh

G c

6pr ot on s

(6pr ot on s ⋅ secon d s) = 1secon d = C

1pr ot on s

⋅

4πh

G c

1pr ot on s

(6pr ot on s ⋅ secon d s) = 6secon d s = H

2pr oton s

⋅

4πh

G c

2pr oton s

(6pr ot on s ⋅ secon d s) = 3secon d s = He

3pr ot on s

⋅

4πh

G c

3pr ot on s

(6pr ot on s ⋅ secon d s) = 2 seco n d s = Li

2pr oton s

⋅

4πh

G c

2pr oton s

(6pr ot on s ⋅ secon d s) = 3secon d s = He

4pr ot on s

⋅

4πh

G c

4pr ot on s

(6pr ot on s ⋅ secon d s) = 1.5secon d = Be

G c

∫

dt = r

3−

= m

G c

∫

dt = (9.2E − 12m /s)

∫

1sec+2sec+3sec

5.52E − 11m

= r

+ r

= 5.6E − 11m + 5.3E − 11m = 1.1E − 10m

= r

+ r

= 5.52E − 11m + 5.5E − 11m = 1.1E − 10m

of 29 53

Oxygen

Where experimentally

And, theoretically

So we are in good agreement.

2−

= m

G c

∫

dt = (9.2E − 12m /s)

∫

1sec+1.5sec+3sec

5.06E − 11m

O = r

+ r

= 4.8E − 11m + 5.3E − 11m = 1.01E − 10m

O = r

+ r

= 5.06E − 11m + 5.5E − 11m = 1.056E − 10m

of 30 53

Part 3: The Solar Formulation For The Solution Of The Wave Equation

of 31 53

The Solar Formulation

Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the

mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the

Sun. In our lunar formulation we had:

We remember the Moon perfectly eclipses the Sun which is to say

Thus equation 1 becomes

The kinetic energy of the Earth is

Putting this in equation 3 gives the mass of the Sun:

We recognize that the orbital velocity of the Moon is

So equation 5 becomes

This gives the mass of the Moon is

Putting this in equation 1 yields

K E

= 3

⊙

⋅

2ℏ

⊙

K E

= 3

⋅

2ℏ

⊙

K E

⋅

⊙

= 3r

⋅

ℏ

⊙

= 3r

⋅ v

⋅

ℏ

⊙

ℏ

⊙

K E

⊙

⋅

⊙

of 32 53

We now multiply through by and we have

10.

Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular

momentum quantized, the angular momentum we will call , the subscript for Planck. We

have

We write for the solution of the Earth/Sun system:

11.

Let us compare this to that of an atom:

12.

We notice that in equation 11

; ; ; ;

is really . We can write 11 as

13.

We say. . That is

Let us see how accurate our equation is:

K E

⊙

⋅

⊙

ℏ

⊙

= r

= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅

= r

= 7.4483E 77J ⋅ m

⋅ kg = 8.3367E 77kg

⋅

K E

⊙

⋅

⊙

E = −

⋅

2ℏ

→

⊙

→ G

→ M

⊙

ℏ

→ L

ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

= ℏ

⊙

/2π

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

⊙

⋅

⊙

2ℏ

⊙

of 33 53

We have that the kinetic energy of the Earth is

Our equation has an accuracy of

Which is very good.

We call the same in our solar solution. But we now want our solution for the solar

formulation. is the mass of the Earth and is the mass of the Sun. We have our solution

might be!

14. !

Where is !

We have

15.

This has an accuracy of

Thus the solutions to the wave equation!

⊙

(6.67408E − 11)

(5.972E 24kg)

(1.9891E 30kg)

2(8.3367E 77kg

⋅

)

⊙

(6.759E 30J )

⊙

6.957E8m

1737400m

= 400.426

K E

= 2.70655E 33J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

2.70655E 33J

2.7396E 33J

= 98.79 %

ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

(9.13E 38)

(6.67408E − 11)(5.972E 24kg)

⋅

400.5986

= 1.4638E11m

1.4638E11m

1.496E11m

100 = 97.85 %

of 34 53

for the solar formulations are!

13)

14)

Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit

Let us equate equation 1 with equation 13:

13.

This gives:

16.

We remember that

−

ℏ

[

∂

∂r

(

∂

∂r

)

sinθ

∂

∂θ

(

sinθ

∂

∂θ

)

sin

∂

∂ϕ

]

ψ + V(r)ψ = E ψ

K E

⊙

⋅

⊙

2ℏ

⊙

ℏ

⊙

⋅

⊙

ℏ

⊙

= 9.13E 38J ⋅ s

⊙

= 2π ℏ

⊙

= 5.7365E 39J ⋅ s

K E

= n

⊙

⋅

2ℏ

⊙

K E

⊙

⋅

⊙

2ℏ

⊙

⋅

2ℏ

⊙

⋅

⊙

⋅ ℏ

⊙

ℏ

⊙

= (hC )K E

hC = 1secon d

C =

⋅

2π r

G m

of 35 53

This gives

17.

We have

18.

This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in

terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,

and the orbital number of the Earth. Let us compute

19.

Let us see how well equation 18 works. at aphelion is 966 m/s and .

. We have

That is an accuracy of

Equation 18 can be written:

20.

From our equation:

6α

h 4π r

= 1.004996352secon d s

6α

⋅

h 4π

⋅

⊙

(1secon d )

⊙

(5.972E 24kg)

(1.9891E 30kg)

(7.34763E 22kg)

(1.732)

= 321,331.459 ≈ 321,331

= 29,800m /s

= 1AU = 1.496E11m

2(966m /s) =

(29,800m /s)

1.496E11m

(1sec)(321,331.459)

1,907m /s = 1,932m /s

1907

1932

= 98.7 %

1secon d = 2r

⊙

6α

h 4π r

= 1.004996352secon d s

of 36 53

We have

21.

Since , the diameter of the Earth orbit, we have

22.

And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice

versa. We basically have

23. 62)

Where is the Earth orbital number. We have

= 7.83436E4seconds

EarthDay=(24)(60)(60)=86400 seconds,

accuracy

This last equation, equation 62, we can use to find the rotation period, or the length of the day,

of an earth-like planet in the habitable zone of any star system, so it would be very useful.

We want to turn our attention to equation 20 and write it

6α

⋅

h 4π

= 2r

⋅

⊙

= d

6α

⋅

h 4π

= d

⋅

⊙

1secon d = 2v

⊙

1secon d =

K E

moon

K E

earth

Ear th Day

1secon d =

Ear th Day

Ear th Day =

⊙

n = 3

Ear th Day =

2(1.496E11m)

966m /s

7.34763E 22kg

1.9891E 30kg

1.732

7.834E4s

86,400s

100 = 90.675 %

of 37 53

24.

We see equating solar and lunar formulations for energy yield the base unit of one second.

Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,

we have

Yields

25.

Where . The accuracy of this is

Thus the energy equations gave the equation 55:

16. !

And equating the orbital equations gives

26.

These last two yield

27.

The accuracy is

1secon d = d

⊙

2ℏ

⊙

⋅

⊙

⋅

⊙

ℏ

⊙

3/2 = cos(π /6)

(2.8314E 33)

(5.972E 24)

(7.34763E 22)

= (0.866)(9.13E 38)

(1737400)

(6.96E8)

4.29059E 72 = 4.5005E 72

4.29059E 72

4.5005E 72

100 = 95.3 %

⊙

⋅ ℏ

⊙

2 3

⋅

⊙

⋅ ℏ

⊙

= 1

of 38 53

Is 98% accuracy. The important thing that comes out of this is our base unit of a second because

we see it is also a function lunar, solar, and earth masses.

24.

(400.5986)

5.972E 24kg

1.9891E 30kg

= 1

0.98 = 1

1secon d = d

⊙

of 39 53

Part 4: Wave Solutions For Jupiter And Saturn

of 40 53

Solutions For Jupiter And Saturn

We have the solution for the Bohr hydrogen atom is

is the atomic number, the number of protons orbited, but for our system it is 1, because one

body is orbited. is the orbit number, which we will deal with later. We have our solution for the

Earth/Moon/Sun system is

Let is the earth mass, is the lunar mass, and is our

Planck constant for the Earth/Moon/Sun system. We compute equation 2:

The kinetic energy of the Earth is

This is

The Moon perfectly eclipses the Sun which means as seen from the Earth the Moon appears to

be the same size as the Sun. This is because

Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,

and is the Moon’s radius. This is because though the Moon is 400 times further from the Sun

than it is from the Earth, the Sun is 400 times larger than the Moon. That is

Thus

We find the quantum mechanical solution to the Earth/Moon/System is

E = −

)

2ℏ

E =

⊙

M = M

m = M

⊙

= 2.8314E 33J ⋅ s

E =

(6.67408E − 11)

(5.972E 24kg)

(7.347673E 22kg)

2(2.8314E 33)

3.93E 30Joules

K E

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

1/n

= 697

≈

⊙

6.96E8m

1737400m

= 400.5986

⊙

= 400

of 41 53

5 !

for Earth orbit (Third Planet).!

We see the solar Planck constant we developed works in solving the Schrödinger wave equation

for the Earth/Moon/Sun system. My belief as to why this was never done is that it had to be

realized that the solution of the Earth kinetic energy around the Sun as a quantum mechanical

state is based on the Moon around the Earth.

Now we want to find what the wave equation solutions are for Jupiter and Saturn because they

significantly carry the majority of the mass of the solar system and thus should embody most

clearly the dynamics of the wave solution to the Solar System.

I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,

and go to the gas giants and ice giants, the atomic number is no longer squared and the square

root of the the orbital number moves from the numerator to the denominator. I believe this is

because the solar system here should be modeled in two parts, just as it is in theories of solar

system formation because there is a force other than just gravity of the Sun at work, which is the

radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on

this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the

radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the

solar system forms, which are gases such as hydrogen and helium, while the heavier elements

are too heavy to be blown out from the inside of the asteroid belt, allowing for the formation of

the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the atomic

number of the heavier metals such as calcium for the Earth, while the equation for the giants has

the atomic numbers of the gasses. We write for these planets

So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):

E = n

⊙

⋅

⊙

n = 3

3(400.5986)(3.93E30J ) = 2.7268585E 33J

2.7268585E33

2.7396E33

= 99.5 %

E =

2ℏ

⊙

K E

(1.89813E 27kg)(13720m /s)

= 1.7865E 35J

E =

(6.67408E − 11)

(1.89813E 27kg)

(7.347673E 22kg)

2(2.8314E 33)

E =

(3.971E 35J ) = Z

(1.776E 35J )

of 42 53

Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that

Our equation for Jupiter is

Where is the atomic number of hydrogen which is 1 proton, and for the orbital

number of Jupiter, . Now we move on to Saturn…

The equation for Saturn is then

It makes sense that Saturn would use Helium in the equation because Saturn is the next planet

after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,

just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.

Our equation in this paper for the Earth orbit does not depend on the Moon’s distance from the

Earth, only its mass. The Moon slows the Earth rotation and this in turn expands the Moon’s

orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by

0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year. It is believed

the Moon came from a chunk knocked off the Earth from a collision with a Mars sized

protoplanet. At the time that the Moon formed it was about 4 Earth radii distant 4.5 billion

years ago. After 100 million years the Moon became tidally locked making its rotation period

equal to its orbital period keeping one face always towards the Earth. After 500 million years the

Moon was orbiting at about 20 Earth radii. It is believed that during the period of heavy

bombardment in a chaotic early solar system about 4.1 to 3.8 billion years ago a large object did

a close pass pulling the Moon further changing its orbit giving it its 5 degree offset from the

Earth’s equator. Our wave equation solution may only use the Moon’s mass but the equation for

1.7865E 35J

1.776E 35J

= 1.006proton s ≈ 1.0proton s = hydrogen(H )

Z = Z

E =

2ℏ

⊙

n = 5

K E

(5.683E 26kg)(10140m /s)

= 2.92E 34J

E =

(6.67408E − 11)

(5.683E 26kg)

(7.347673E 22)

2(2.8314E 33)

2.45

(3.5588E 34J ) = Z(1.45259E 34J )

Z(1.45259E 34J ) = (2.92E 34J )

Z = 2protons = Helium(He)

E =

2ℏ

⊙

of 43 53

kinetic energy of the Moon to kinetic energy of the Earth times the Earth Day equal to about one

second:

Which we connect with the equation where the proton gives one second

This holds for when the Moon was at a distance from the Earth such that it appears the same

size as the Sun, which means:

Which is when the two equations above for one second connect to our wave equation solution to

the Earth

Because .

The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the

Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get too

extremely hot or too extremely cold. We see the Moon may be there for a reason.

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

6α

h 4π r

= 1.004996354secon d s

planet

moon

≈

star

moon

K E

= n

⊙

⋅

2ℏ

⊙

star

moon

⊙

of 44 53

Part 5: Modeling Habitable Star Systems

of 45 53

Modeling Habitable Star Systems

We want to solve a habitable planetary system in general and apply it to another star system. We

must include the solution of its moon as well, because the Moon of the Earth makes life

optimally possible. However, our Moon is counter-intuitive, it breaks all of the rules of the other

moons in our solar system. One is it has a low mass for its size. It has been said that it could be

hollow. Even suggested that it is a hollow-spacecraft put there to make life as we know it

possible. In our quantum theory for the solar system, we may have discovered the science of the

engineers who may have made the Moon, and we are possibly applying their science here. Our

quantum theory for the solar system also has a solar solution as compared to a lunar solution, so

it works for star systems in general.

The perfect star system to use for our purposes here would be Tau Ceti. Tau Ceti is a star

spectrally similar to the Sun and the closest of such a G-class star, at about 12 light years distant.

Its mass is 78% of the Sun’s. Its mass is

Its radius is:

And its luminosity is

It has several unconfirmed planets with one potentially in the habitable zone. It has been the

subject of science fiction literature. It is in the constellation Cetus. From Tau Ceti the Earth

would be seen to be in the northern hemisphere constellation Bootes with an apparent

magnitude of 2.6. It is spectral class G8 V where our Sun is G2 V.

For the luminosity of a main sequence star we only need to know the mass of the star to get its

luminosity:

This holds in solar masses and solar luminosities. The habitable zone of a star is given by

luminosity. If the star is 100 times brighter than our Sun, then the habitable zone is 10 times

further than the Earth is from the Sun by the inverse square law.

In astronomical units and solar luminosities. From we can get the orbital period in Earth

years:

We can get the radius of the Moon of the planet if we know the mass of the planet, :

= 0.783 + / − 0.012M

⊙

= 0.793 + / − 0.004R

⊙

= 0.488 + / − 0.0010L

⊙

L = M

3.5

= L

= r

of 46 53

We want the mass of the Earth from the mass of the Sun. We would guess we can get it from the

mass of the Sun, the size of the Sun, and the base 60 sexagesimal system upon which we have

found everything is based. I find the relationship exists for the Earth:

(36)(5)=180=360/2 and 360/6=60 is base 60 sexagesimal. This gives

This is an accuracy of

We can get the orbital distance of the planet’s Moon from the planet:

Once we have that we can get the radius of the Moon of the planet

Now we can get the rotation period of the planet, its length of day:

Because the orbital velocity of the planet’s Moon is

We can use the the delocalization time get the moon’s mass, which is 1/2 the planet’s year, with

10 , See Appendix 1

⊙

= 1

⊙

= 0.13889(1.9891E 30kg)

6.96E8m

1.496E11m

= 5.979748E 24kg

5.972E 24

5.979748E 24

100 = 99.87 %

planet

moon

≈

star

moon

⊙

= 1

Ear th Day =

⊙

τ =

moon

(2r

moon

)

ℏ

of 47 53

And the whole system is solved. Let us see how the theoretical luminosity appears with that

measured with equation 8.1:

It is close to the measured. The equation for predicting is approximate as the actual

values can vary a little with things like metallicity of the stars. Tau Ceti has low metallicity

compared to our Sun.

Let’s find the orbital distance of the habitable planet if it exists with equation 2:

Let’s find the orbital period of the habitable planet (It’s year) with equation 3:

Let’s get the mass of the habitable planet from equation 5:

Which is good because the planet, if it exists, is thought to be larger than the Earth. Let’s get the

radius of its moon from equation 4:

Let’s see at what distance it orbits the planet with equation 6:

= M

3.5

= 0.783

3.5

= 0.425L

⊙

0.448L

⊙

= 0.488 = 0.69857AU = (0.6986AU )(1.496E11m /AU ) = 1.045E11meters

= r

= (0.69867AU )

= 0.584years(365.25d a ys)((24hrs)(60min)(60sec) = 1849638secon d s

(0.584)(365.25d ays) = 213.306d a ys

= (0.783)(1.9891E 30kg) = 1.5575E30kg

= (0.793)(6.96E8m) = 5.52E 8m

(1.5575E 30kg)

(5.52E8m)

(1.045E11m)

= 6.036E 24kg

6.036E 24

5.972E 24

= 1.0107Ear th Ma sses

= 2R

2(5.52E 8m)

(6.036E 24kg)

(1.5575E 30kg)

= 1.53656E6m =

1.53656E6m

1.7374E6m

= 0.8844Lu n arRa dii

of 48 53

Let’s get the orbital velocity of this moon with equation 9:

So the orbital period of the moon is:

Compared to that of the Earth which is 27.3 days for the sidereal month and is a 29.53day lunar

month. The Moon of Tau Ceti would have a slightly longer lunar month, too. Now we can

determine the rotation period of the planet which would be its day:

We get the mass of the moon from equation 10:

So the rotation period of the planet (Its Day) is from equation 8:

If the moon of Tau Ceti is to perfectly eclipse its star, like our Moon does with the Sun, to let its

inhabitants know that they are there for reason, and so as to theoretically be a condition for

optimizing life, then we have the following should hold:

moon

= r

planet

moon

star

= (1.045E11m)

1.53656E6m

5.52E8m

= 2.91E8m

2.91E8m

3.84E8m

= 0.7578L u n ar OrbitalRa dii

(6.67408E − 11)(6.0636E 24kg

2.91E8m

= 1,176.6m /s

2π r

2π (.91E 8m)

1,176.6m /s

= 1.554E6secon d s

(1.554E6s)

min

60sec

h our

60min

d ay

24h ours

= 17.987d ays

ℏ

(2.8314E 33J ⋅ s)(0.5)(18429638s)

4(2.91E 8m)

= 7.7E 22kg

7.7E 22kg

7.34763E 22kg

= 1.048Ear th Moon s

Pl a n et Day =

2(1.045E11m)

1,176.6m /s

7.7E 22kg

1.5575E 30kg

= 39495.61secon d s = 658.26min = 10.971h ours

10.971

24h ours

= 0.457Ear th Days

planet

moon

≈

star

moon

of 49 53

We have

And it does hold. For the Earth/Moon/Sun system we get

For the Tau Ceti system the 359 is close to 360, which is a convenient amount of degrees into

which divide a circle, like we did. We see the Tau Cetians might do the same, because like us

they might choose the base 60 sexagesimal system of counting, and this 360 might influence the

way the inhabitants of a planet orbiting this star would make their calendar. Just like the 365

day year here on Earth corresponds closely to the 360 degrees of a circle. Aside from having here

on Earth the degree, we have gradians, of which there are 400 in a circle. They make it very easy

for computing right angles to a given angle. They are used more in Europe and in surveying and

architecture, though it might have some interesting connection with our 400=400.

The habitable planet of Tau Ceti would orbit it star with a period of 213 days, which would be its

year. It star is about 78% the mass of our Sun, and about 79% its size. Its habitable planet would

be a little larger than the Earth, but not much. Its Moon would be about 88% the size of our

moon, but a little bit more massive, but not much more massive. It would orbit Tau Ceti once

about every 18 days meaning there would be 11.833 months in the Tau Ceti Year, or about 12

months like we have. It would orbit the planet at about three quarters the distance ours does.

The day from sunrise to sunrise, or sunset to sunset would be about 11 hours, or close to half the

length of our day. This is if Tau Ceti was engineered for life, but it doesn’t have to have a planet

so optimally designed for life as ours. Our 24 hour day would be better for when entering the

realm of doing astronomy, because longer nights mean more complete observations of the

universe surrounding us than for the Tau Cetians. It may mean for the conditions for life to be

optimal, like here on Earth, the star would have to be as massive as the Sun, to provide a longer

year and a longer day. But we see here how the science of engineering planetary systems optimal

for life, might be done.

1.045E11m

2.91E8m

5.52E8m

1.53656E6m

359 = 359

400 = 400

of 50 53

Appendix 1

If we want to prove that our planetary Planck constant is correct, the delocalization time for the

Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We want

to solve the Schrödinger wave equation for a wave packet and use the most basic thing we can

which is a Gaussian distribution. We want to then substitute for Planck’s constant that is used

for quanta and atoms our Planck-type constant (h bar solar) for the Earth/Moon/Sun system

then apply it to predict the delocalization time for the Moon in its orbit with the Earth around

the Sun.

We consider a Gaussian wave-packet at t=0:

We say that is the delocalization length and decompose the wave packet with a Fourier

transform:

is the harmonics of the wave function. We use the identity that gives the integral of a

quadratic:

Solve the equation

With the initial condition

A plane wave is the solution:

Where,

The wave-packet evolves with time as

Calculate the Gaussian integral of

ℏ

⊙

ψ (x,0) = Ae

−

ψ (x,0) = Ae

−

∫

2π ℏ

ℏ

∫

∞

−∞

−α

x+βx

d x =

4α

iℏ∂

ψ (x, t) =



ψ (x, t)

ψ (x,0) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−ϵ( p)t)

ϵ( p) =

ψ (x, t) =

∫

dp ⋅ e

2ℏ

⋅ e

−

ℏ

( px−

of 51 53

and

The solution is:

where

is the delocalization distance, which for instance could be the width of an atom. is the

delocalization time, the average time for say an electron to traverse the diameter of the atom and

even leave it, to delocalize. If we substitute for our , and say that the delocalization distance

is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization

time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We

have

Where is the mass of the Moon, and is the orbital radius of the Moon. We have

Now let’s compute a half a year…

(1/2)(365.25)(24)(60)(60)=15778800 seconds

So we see our delocalization time is very close to the half year over which the Earth and Moon

travel from one position to the opposite side of the Sun. The closeness is

Thus we know our is accurate, it continues to function in a theoretical framework. The thing

about this is that it means we can predict the mass of the Moon from the Earth year. In terms of

what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at

its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a

mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for

water to exist as liquid, and thus could be as it is for a reason, which means life might be part of

a physical process throughout the Universe, that it unfolds naturally in the evolution of star

systems.

2ℏ

2mℏ

β =

i x

ℏ

= ex p

[

−

⋅

1 + t

/τ

]

τ =

m d

ℏ

⊙

τ =

moon

(2r

moon

)

ℏ

⊙

moon

τ = 4

(7.34767E 22kg)(3.844E8m)

2.8314E 33J ⋅ s

= 15338227secon d s

15338227

15778800

100 = 97.2 %

ℏ

⊙

of 52 53

Appendix 2: The Data For Verifying The Equations

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and

Earth’s orbital velocity at perihelion we have:

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G: 6.67408 × 10

−11

c : 299,792,458m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

K E

moon

(7.347673E 22kg)(966m /s)

= 3.428E 28J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

of 53 53

The Author!